题目内容
将4.6g钠投入95.4g水中,所得到溶液的质量分数是( )
A.等于4.6% | B.等于8% | C.大于8% | D.小于8% |
将4.6g钠投入到95.4g水中,发生2Na+2H2O=2NaOH+H2↑,则
2Na+2H2O=2NaOH+H2↑
46g 80g 2g
4.6g m(NaOH)m(H2)
m(NaOH)=
=8.0g,
m(H2)=
=0.2g,
则w(NaOH)=
×100%=8.02%>8%,
故选D.
2Na+2H2O=2NaOH+H2↑
46g 80g 2g
4.6g m(NaOH)m(H2)
m(NaOH)=
4.6g×80g |
46g |
m(H2)=
4.6g×2g |
46g |
则w(NaOH)=
8.0g |
(4.6g+95.4g-0.2g) |
故选D.
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